Timeline for Sums of two squares in (certain) integral domains
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24 events
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| Apr 27, 2022 at 14:30 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 27, 2022 at 14:21 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Jul 10, 2010 at 19:04 | vote | accept | Pete L. Clark | ||
| Jul 10, 2010 at 19:04 | vote | accept | Pete L. Clark | ||
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| Jul 8, 2010 at 17:49 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jul 8, 2010 at 9:21 | answer | added | Franz Lemmermeyer | timeline score: 9 | |
| Jul 8, 2010 at 7:41 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jul 8, 2010 at 6:52 | comment | added | Pete L. Clark | @KConrad -- yes, I understand what you are saying, and eventually I'll look in F&T to see if it's actually impossible. (Good thing I remembered the ring of functions on the unit circle example; otherwise I'd be trying to prove that such domains R do not exist.) @KBuzzard -- yes, that's exactly the trap I fell into. In MAGMA's defense, I was doing this calculation very quickly towards the end of a long day, and I didn't stop to check up on what I was doing -- as soon as I saw discriminant = discriminant, I claimed victory. | |
| Jul 8, 2010 at 6:49 | comment | added | KConrad | In my previous comment, the Z[sqrt{5}] should be Z[sqrt{-5}]. | |
| Jul 8, 2010 at 6:43 | comment | added | Kevin Buzzard | @Pete: if $K$ is a number field in magma then Discriminant(K) does not compute the discriminant of $K$---it computes the discriminant of the polynomial used to define $K$. I am wondering whether you fell into this awful trap and you wouldn't be the first. To compute disc(K) in magma you need to compute the integer ring explicitly and then compute the discriminant of that. | |
| Jul 8, 2010 at 6:34 | comment | added | KConrad | Last comment for now: to avoid any misunderstandings, I'm not saying you can't get biquadratic fields Q(sqrt(d),i) with class number 1 when Q(sqrt(d)) has class number greater than 1. The field Q(sqrt(−5),i) has class number 1 and Z[sqrt{5}] has class number 2. The problem is to also rig the example so the ring of integers is R[i] where R is the ring of integers of Q(sqrt(d)). That is not seeming easy to add to the mix. | |
| Jul 8, 2010 at 6:25 | comment | added | KConrad | Oh, about my "pretty common" comment involving class number 1. I didn't mean any rigorous theorem, but just that random number fields often seem to have class number 1 (unless you carefully choose your setting, e.g., imaginary quadratic). I have tried picking some cubic and quartic fields with class number greater than 1 using some families of polynomials and anytime I get one with $h > 1$, adjoining $i$ is producing a number field with class number greater than 1. Too bad for you. | |
| Jul 8, 2010 at 6:11 | comment | added | KConrad | By "easier to get examples" I meant "easier to get examples where ${\mathbf Q}(\sqrt{d})$ has class number greater than 1", but invariably they are not leading to examples which you want, where adjoining $i$ gives a number field with class number 1. | |
| Jul 8, 2010 at 6:08 | comment | added | KConrad | I did some checking with PARI and cases where ${\mathbf Q}(\sqrt{d})$ has class number greater than 1 with $d \equiv 1 \bmod 4$ are not leading to ${\mathbf Q})(\sqrt{d},i)$ with class number 1 for small $d$ (easier to get examples if $d < 0$ but I looked at $d > 0$ too). There are Brauer relations describing the class number of the biquadratic field in terms of class numbers of quadratic subfields (see Frohlich--Taylor's book at the end). Probably they will confirm this numerology happens all the time, in which case you won't get examples with $R$ the integers of a quadratic field. | |
| Jul 8, 2010 at 6:07 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jul 8, 2010 at 6:05 | comment | added | Pete L. Clark | In other news, a MATHSCINET search turns up the 1980 paper of Choi, Lam, Reznick and Rosenberg, which seems to contain closely related results. For them, both $R$ and $R[i]$ are UFDs. (I don't have online access to the paper, unfortunately.) So the most interesting aspect again seems to be the situation in which $R$ is not a UFD. | |
| Jul 8, 2010 at 6:02 | comment | added | Pete L. Clark | @K: It seems like your first two comments are sort of at odds with each other: first, maybe you can help me get an explicit example of a quadratic field whose ring of integers $R$ satisfies this property, since the one I mentioned doesn't work. Also, by "lots" of number fields, do you mean "infinitely many"? Not provably, I'm sure. So then what? | |
| Jul 8, 2010 at 5:59 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jul 8, 2010 at 5:58 | comment | added | Pete L. Clark | @K: Hmm, I'm sure you're right. I worried about this point but thought I was OK since MAGMA told me that the discriminant of the number field $\mathbb{Q}(\sqrt{-5},\sqrt{-1})$ was equal to the discriminant of the minimal polynomial of $\sqrt{-1} + \sqrt{-5}$. I will retract the 4th corollary for now, although probably it can be fixed by inverting $2$... | |
| Jul 8, 2010 at 5:57 | comment | added | KConrad | Clarify what you mean in the first question about having seen anything "like this result" before. Do you mean the specific application to Artin--Pfister? The proof is "follow your nose" as you say, copying the standard technique for deciding how a prime factors in a ring generated by a root of a polynomial by seeing how the polynomial with that root factors mod the prime (assuming the ring is generated by the root!). As for the second question, this "strange situation" is pretty common since lots of number fields with class number > 1 are inside a number field with class number 1. | |
| Jul 8, 2010 at 5:50 | comment | added | KConrad | Pete, $R = {\mathbf Z}[\sqrt(-5)]$ is not an example since the ring of integers of ${\mathbf Q}(\sqrt(-5),i)$ is not $R[i]$ but something bigger: the ratio $(i + sqrt(-5))/2$ is an algebraic integer. The ring of integers in fact is ${\mathbf Z}[(i+\sqrt(-5))/2]$. If $K$ is a quadratic field unramified at 2 then the ring of integers of $K(i)$ is $R[i]$ where $R$ is the ring of integers of $K$. So you need $K = {\mathbf Q}(\sqrt{d})$ with $d \equiv 1 \bmod 4$. I don't believe the 4th corollary now except perhaps if you allow 1/2's, but I don't have a specific counterexample, say for $p=3$. | |
| Jul 8, 2010 at 5:40 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jul 8, 2010 at 5:32 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jul 8, 2010 at 5:26 | history | asked | Pete L. Clark | CC BY-SA 2.5 |